Answer : The rate constant at 785.0 K is,
Explanation :
According to the Arrhenius equation,
or,
where,
= rate constant at =
= rate constant at = ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature =
Now put all the given values in this formula, we get:
Therefore, the rate constant at 785.0 K is,
Answer:
The answer to your question is: C. The specific latent heat of fusion
Explanation:
A. The specific latent heat of vaporization Specific latent heat of vaporization indicates the transition from liquid to vapor, but we are not looking for this definition. This answer is wrong.
B. The specific heat
indicates the amount of heat needed to increase the temperature of water 1°C, so this answer is wrong.
C. The specific latent heat of fusion
. This heat indicate the transition from solid ie to liquid, so this is the right answer.
D. The internal energy measures the energy of the molecules of a substance, so this answer is wrong.
Answer:
The answer is an attached file
Explanation:
I hope it'll be useful to you.
Answer:
Even the most powerful light-focusing microscopes can't visualise single atoms. What makes an object visible is the way it deflects visible light waves. Atoms are so much smaller than the wavelength of visible light that the two don't really interact. To put it another way, atoms are invisible to light itself.
Explanation:
Can you give me Brainliest pls
And Your welcome :)
The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.
The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.
The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.
Thorium has several isotopes, with different rates and types of emission:
- ²³²Th emits α particles, it is the most abundant 99.9%
- ²³⁴Th emits β particles, exists in small traces.
In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.
Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.
In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
Learn more about radioactive emission here: brainly.com/question/15176980